Two different comultiplications on a Hopf algebra

Question

I am pretty sure this statement is false : let $K$ be a field and let $(A, \eta, \mu, \Delta, \epsilon, c)$ be a Hopf Algebra. If we forget the comultiplication $\Delta$, is it forced by $(A, \eta, \mu, \epsilon, c)$ ? In other words, can we put two Hopf algebra structures on a $K$-algebra which only differ by their comultiplication ?

Answer 1

The answer to your second question is, in principle, yes.

Here's my argument: Let us take a Hopf algebra $(H,m,\eta,\Delta,\varepsilon,S)$. Let us suppose that it is commutative (as an algebra) but not cocommutative (as a coalgebra). So $\Delta\neq\tau\circ\Delta$. On the other hand, commutativity implies that $S^2=Id \ $, i.e. $\ S=S^{-1}$.

Let us now define the Hopf algebra $H^{cop}$, which is the same underlying space, with the same ope-rations, except the comultiplication, which is now defined by $\Delta^{cop}=\tau\circ\Delta$, with $\tau$ standing for the twist map $$\tau:H\otimes H\rightarrow H\otimes H, \ \ \ \tau(a\otimes b)=b\otimes a$$ We can now show that $H^{cop}$ is also a Hopf algebra. Now, $H\equiv (H,m,\eta,\Delta,\varepsilon,S)$ and $H^{cop}\equiv(H,m,\eta,\Delta^{cop},\varepsilon,S)$ have the same underlying spaces and operations except that they differ in the comultiplication.

An example of a Hopf algebra, satisfying the above requirements (commutative but not cocommutative), is the dual Hopf algebra $(kG)^*$ of the group Hopf algebra $kG$, where $G$ is a finite, non-abelian group.